Stationary States

What makes a state "stationary"?

When the Hamiltonian $\hat{H}$ does not depend on time, a special class of solutions to the time-dependent Schrödinger equation (TDSE) exists: solutions where the spatial and time parts factorize. Writing $\Psi(x, t) = \psi(x)\, T(t)$ and substituting into

$i\hbar \frac{\partial \Psi}{\partial t} = \hat{H}\Psi,$

the left side contributes $i\hbar\, \psi(x)\, T'(t)$ and the right side gives $T(t)\,\hat{H}\psi(x)$ . Dividing through by $\psi(x) T(t)$ separates the equation into two independent sides: one depending only on $t$ , the other only on $x$ . Each side must therefore equal the same constant, which we call $E$ :

$i\hbar \frac{T'(t)}{T(t)} = E \qquad \text{and} \qquad \hat{H}\psi(x) = E\,\psi(x).$

The second equation is the time-independent Schrödinger equation (TISE); solutions $\psi(x)$ are called energy eigenstates (or stationary states) with associated energy eigenvalue $E$ .

The time phase factor

The time equation $i\hbar\, T'(t) = E\, T(t)$ is an ordinary first-order ODE with solution

$T(t) = e^{-i E t / \hbar} = e^{-i \omega t},$

where $\omega \equiv E / \hbar$ is the angular frequency of the phase oscillation.

The full solution for an energy eigenstate is therefore

$\Psi_n(x, t) = \psi_n(x)\, e^{-i E_n t / \hbar}.$

Why "stationary"?

Even though the wave function picks up a time-dependent phase, every observable probability density is completely static:

$|\Psi_n(x, t)|^2 = |\psi_n(x)|^2\, |e^{-i E_n t / \hbar}|^2 = |\psi_n(x)|^2.$

The modulus of a complex exponential $e^{i\theta}$ is always 1 for real $\theta$ , so $|e^{-i\omega t}|^2 = 1$ . The spatial probability density $|\psi_n(x)|^2$ is frozen in time — nothing observable about the state changes, which is exactly why these solutions earn the name "stationary."

The same argument applies to expectation values: any observable $\hat{Q}$ that does not explicitly depend on time satisfies $\langle \hat{Q} \rangle = \text{const}$ for a stationary state, because the phase factor cancels between the bra and the ket.

The angular frequency in numbers

The oscillation frequency $\omega = E/\hbar$ sets the "tick rate" of the phase. Because $\hbar$ is tiny ( $1.055 \times 10^{-34}$ J $\cdot$ s), even a modest energy of $1$ eV ( $= 1.602 \times 10^{-19}$ J) produces a very high frequency:

\approx 1.52 \times 10^{15}\,\text{rad/s}.$$ That frequency lies in the near-infrared (a $1$ eV quantum corresponds to a wavelength of about $1240$ nm) — a reminder that eV-scale energies sit in the optical region of the spectrum. Visible light spans roughly $1.6$–$3.3$ eV, so atomic energy scales of a few eV correspond to the infrared-through-near-UV light seen in atomic emission spectra. ## Try it This is a numerical exercise — `return` a number. A quantum particle is in a stationary state with energy $E = 1.0\,\text{eV}$. What is the angular frequency $\omega = E/\hbar$ of its time phase factor $e^{-i\omega t}$ (in rad/s)? Use $1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J}$ and $\hbar = 1.055 \times 10^{-34}\,\text{J·s}$. <RunnableExample />